IIT - JEE 2015 (Advanced) - Vidyalankarvidyalankar.org/file/JEE-Advanced-2015-Solution-Paper-I.pdf · (4) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution 4 6. A Young’s

IIT - JEE 2015 (Advanced)

CODE : 8

(2) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution

2

PART I : PHYSICS

SECTION 1 (Maximum Marks : 32)

This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened 0 In all other cases 1. Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation

of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm)

1. [2]

n 2

13.6E eV

n 2 2

1 23 3

g 30 g 274 4

Energy of photon = phc 1242 eV nm 138

E eV 13.8 eV90 nm 10

K = 10.4 eV [Recoil energy of atom is negligible]

2

13.6 eV13.8 eV 10.4 eV

n

[By conservation of Energy]

2 13.6n 4 n 2

3.4

2. A bullet is fired vertically upwards with velocity v from the surface of a spherical planet.

When it reaches its maximum height, its acceleration due to the planets gravity is ¼th of

its value at the surface of the planet. If the escape velocity from the planet is vesc = v N , then the value of N is (ignore energy loss due to atmosphere)

2. [2]

x s 2 2

1 1 1g g x 2R

4 x 4R [R = Radius of planet]

2GM 1 GMmV

R 2 2R 2 GM

VR

2e

2GMV N 2

R

3. Two identical uniform discs roll without slipping on two different surfaces AB and CD

(see figure) starting at A and C with linear speeds v1 and v2, respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed v1 = 3 m/s, then v2 in m/s is (g = 10 m/s2)

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – I) (3)

3

3. [7] Final kinetic energies are equal

2 21 2

3 3v g 30 v g 27

4 4

2 22 1v v 4g 9 40

12v 7ms

4. Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that

of B and A emits 104 times the power emitted from B. The ratio A

B

of their

wavelengths A and B at which the peaks occur in their respective radiation curves is 4. [2]

1/44 2 4

A A A A A B A B4 2 4

B A B AB B B B

P A T R T T R P

P T R PA T R T

By Wein’s law : 1/4

A B4

B A

T 1400 2

T 10

5. A nuclear power plant supplying electrical power to a village uses a radioactive material

of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is

5. [3] Power generated Activity Requirement = x Initial activity = A0

x = 0.125 A0 after time ti activity becomes = x

t0 0

10.125A A e t ln 0.125

T 1

ln 2

ln8

t T 3Tln


4

6. A Young’s double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by x2 = p2m2

2 d2, where is the wavelength of light in air (refractive

index = 1), 2d is the separation between the slits and m is an integer. The value of p is

6. [3]

Optical path length S2P = 2 2x d [P is point of interference on water surface]

Optical path length S1P = 2 2x d

2 2x d 1 m , = 4/3 2 2 2 2x 9m d

7. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10

cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M1. When the set-up is kept in a medium of

refractive index 7/6, the magnification becomes M2. Then magnitude 2

1

M

M is

7. [7] Image formation by concave mirror is unaffected by presence of medium.

M

1 1

v 15 =

1

10

M

1

v=

1 1

15 10 =

1

30

vM = 30

in air : 1

f=

1

10=

31 G

2

G =1

5

2u 15

20 cm 140

vm va vm


5

in medium 7/6 : 2

1

f=

3 61 G

2 7

=2

35

in air: a

1 1

v 20

=

1

10=

1

v=

1 1

10 20 =

1

20= va = 20

in medium : m

1 1

v 20

=

2

35=

m

1

v=

2 1

35 20 =

8 7

140

=

m

1

v=

1

400

2

1

M

M= m

a

v

v = 7

8. An infinitely long uniform line charge distribution of charge per unit length lies parallel

to the y-axis in the y-z plane at z = 3

2 (see figure). If the magnitude of the flux of the

electric field through the rectangular surface ABCD lying in the x-y plane with its centre

at the origin is 00

L( permittivity of free space),

n

then the value of n is

8. [6]

2

2

0

30E x

2 4

flux through shaded patch:

d =2

0 2

1

2 3ax

4

2

2

3 a / 2

3ax

4

Lx

E sin

dx

x

3a

2


6

d = 20 2

x3 aL / 2

2 3ax

2

d =a/2

20 a/2

L 1 dx

2 3a x1

2 3a / 2

=1/ 3

20 1/ 3

L dy

2 y 1

= 1

0 0

L 1 L2tan

2 63


This section contains TEN questions Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE

of these four option (s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the

ORS Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened 2 In all other cases

9. Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2. The distance d is

(A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm

9. (B) S1

2 1 2 1

v u R

31

1 3 2v 2 50 10

1 3 1

v 100 20

S2

2 1 2 1

v u R


7

313 1 2

200 d 50 10

1 1

d 50 20

d 50 20 d 70 cm 10. A conductor (show in the figure) carrying constant current I is kept in the x-y plane in a

uniform magnetic field B . If F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is(are)

(A) If B is along z , F (L + R) (B) If B is along x , F = 0

(C) If B is along y , F (L + R) (D) If B is along z , F = 0

10. (A), (B), (C) To find magnetic force on a current wire vector length of current wire is taken. So length

of wire = 2 L + R

and F I L B correct options are (A), (B) and (C). 11. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium

in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is(are) (A) The average energy per mole of the gas mixture is 2RT

(B) The ratio of speed of sound in the gas mixture to that in helium gas is 6 5

(C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2

(D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1 2

11. (A), (B), (D)

1 21 2

n nU f RT f RT

2 2

1 1

5RT 3RT2 2

4RT

Average energy per mole of the mixture = 4RT

2 = 2RT ...(A)

rms

3RTv

M vrms

1

M

2

rms He

rms H

1v 14v 1 2

2

…(D)


8

12. In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 108 m and 1.0 107 m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

(A) 2475

64 (B)

1875

64 (C)

1875

49 (D)

2475

132

12. (C)

RA = A

= 8 3

56

2.7 10 50 103 10

(49 4) 10

7 3

5Fe 6

1.0 10 50 10R 125 10

4 10

5 5

eq A Fe 5 5

3 10 125 10R R || R

3 10 125 10

= 5 63 12510 10

128

= 1875

64

13. For photo-electric effect with incident photon wavelength , the stopping potential is V0. Identify the correct variation(s) of V0 with and 1/. (A) (B) (C) (D)

13. (A), (C) Stopping potential is given by

0hc

Ve e


9

14. Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then: (A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm. (B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm. (C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm. (D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

14. (B), (C)

Smallest division on main scale = 1

8cm = 0.125 cm

5 divisions of the vernier scale = 4 divisions of the main scale = 4 0.125 = 0.5 cm

1 division of vernier scale = 0.5

5 = 0.1 cm

Least count of vernier = 1 main scale division 1 vernier scale division = 0.125 0.1 = 0.025 cm

Least count of screw gauge = pitch

100

If pitch of the screw gauge is twice the least count of vernier calipers then the least count

of screw gauge = 0.05 0.5

cm mm100 100

= 0.005 mm

Hence (B) is correct. Also, least count of linear scale of screw gauge = 2 0.025 cm = 0.05 cm pitch = 2 0.05 cm = 0.1 cm = 1 mm

Least Count = 1mm

0.01mm100

Hence (C) is correct. 15. Planck's constant h, speed of light c and gravitational constant G are used to form a unit of

length L and a unit of mass M. Then the correct option(s) is(are)

(A) M c (B) M G (C) L h (D) L G 15. (A), (C), (D) Let L hx cy Gz L (ML2T1)x (LT1)y (M1L3T2)z L Mxz L2x+y+3z Txy2z x z = 0 …………… (1) 2x + y + 3z = 1 …………… (2) x + y + 1z = 0 …………… (3)


10

On solving

x = 1 3 1

, y , z2 2 2

Option (C) and (D) are correct. Similarly Let M hx cy Gz M Mxz L2x+y+3z Txy2z x z = 1 …………… (1) 2x + y + 3z …………… (2) x + y + 2z = 0 …………… (3) Equation (2) Equation (3) gives x + z = 0 2x = 1

x = 1

2 z =

1

2 y =

1

2

M h , m c Option (A) is correct. 16. Two independent harmonic oscillators of equal mass are oscillating about the origin with

angular frequencies 1 and 2 and have total energies E1 and E2, respectively. Then

variations of their momenta p with positions x are shown in the figures. If a

b = n2 and

a

R = n, then the correct equation(s) is(are)

(A) E11 = E22 (B) 2

1

= n2 (C) 12 = n2 (D) 1

1

E

= 2

2

E

16. (B), (D)

2

2 2 21 1

1

1 b a 1E m a n

2 2m b m

…(i)

2

2 22 2 2

1 RE m R m 1

2 2m …(ii)

From (i) and (ii) 22

1n

2 2

21 1 12

2 2 2

E a 1n

E R n

1 2

1 2

E E


11

17. A ring of mass M and radius R is rotating with angular speed about a fixed vertical axis

passing through its centre O with two point masses each of mass M

8at rest at O. These

masses can move radially outwards along two massless rods fixed on the ring as shown in

the figure. At some instant the angular speed of the system is 8

9 and one of the masses is

at a distance of 3

5R from O. At this instant the distance of the other mass from O is

(A) 2

3R (B)

1

3R (C)

3

5R (D)

4

5R

17. (D) Let the other mass be at a distance x from the centre. Conserving angular momentum

about the axis :

MR2 =

22 2M 3 M 8

MR R x8 5 8 9

on solving x = 4

5R.

18. The figures below depict two situations in which two infinitely long static line charges of

constant positive line charge density are kept parallel to each other. In their resulting electric field, point charges q and q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is (are)

(A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge q continues moving in the

direction of its displacement. (D) Charge q executes simple harmonic motion while charge +q continues moving in the

direction of its displacement.


12

18. (C) The charge experiences resting force whereas ve charge experiences force in the

direction of displacement.

Section III SECTION 3 (Maximum Marks : 16)

This section contains TWO questions Each question contains two columns, Column I and Column II Column I has four entries (A), (B), (C) and (D) Column II has five entries (P), (Q), (R), (S) and (T) Match the entries in Column I with the entries in Column II One or more entries in Column I may match with one or more entries in Column II The ORS contains a 4 5 matrix whose layout will be similar to the one shown below : For each entry in Column I, darken the bubbles of all the matching entries. For example,

if entry (A) in Column I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D).

Marking scheme : For each entry in Column I. +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened 0 If none of the bubbles is darkened 1 In all other cases 19. Match the nuclear processes given in column I with the appropriate option(s) in column (II).

Column I Column II (A) Nuclear fusion (P) Absorption of thermal neutrons by 235

92 U

(B) Fission in a nuclear reactor (Q) 6027 Co nucleus

(C) -decay (R) Energy production in stars via hydrogen conversion to helium

(D) -ray emission (S) Heavy water (T) Neutrino emission

19. (A) (R); (B) (P), (S); (C) (Q), (T); (D) (P), (Q), (R)

(P) (Q) (R) (S) (T) (A)

(P) (Q) (R) (S) (T) (B)

(P) (Q) (R) (S) (T) (C)

(P) (Q) (R) (S) (T) (D)

E E

+q x

E E

q x


13

20. A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I ( and U0 are constants). Match the potential energies in column I to the corresponding statement(s) in column II.

Column I Column II

(A) 220

1U x

U (x) 12 a

(P) The force acting on the particle is zero at

x = a.

(B) 20

2U x

U (x)2 a

(Q) The force acting on the particle is zero at

x = 0

(C) 2 20

3U x x

U (x) exp2 a a

(R) The force acting on the particle is zero at

x = a

(D) 30

4U x 1 x

U (x)2 a 3 a

(S) The particle experiences an attractive force towards x = 0 in the region |x| < a.

(T) The particle with total energy 0U

4 can

oscillate about the point x = a.

20. (A) (P), (Q), (R), (T ); (B) (Q), (S); (C) (P), (Q), (R), (S); (D) (P), (R)

(A)

220U x

U 12 a

2

02

UdU x 2xF 2 1

dx 2 a a

2

02

U x 2x1

2 a a

F = 0 if x = 0

Also, F = 0

if 2x

1 0a

2x

1a

x

1a

x = a


14

(A) P, Q, R

(B) 202

UU x

2a

0 02 2

U UdUF 2x x

dx 2a a F = 0 if x = 0

(C) 2 2

0U x xU exp

2 a a

2x2 2a0 0

2 2

U UdU x 2x x 2xF exp exp

dx 2 a a 2a a

2 2x x2a a0

2

U 2x xe e

2 aa

2x2a

0 2

x xU 1 e

aa

2x 2a0

2

U xx e 1

aa

F = 0 at x = 0, x = a. Also F is negative for |x| < a

P, Q, R, S.

(D) 3

0U x 1 xU

2 a 3 a

2

03

UdU 1 1 3xF

dx 2 a 3 a

2

02

U x1

2a a

F = 0 at x = a. Also F is negative for |x| < a.

P, R, S.


15

PART II : CHEMISTRY


This section contains EIGHT questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened 0 In all other cases 21. The number of resonance structures for N is

21. [7]

O H

2H OOH

O

O

O

O

O

OO

O

22. The total number of lone pairs of electrons in N2O3 is 22. [8]

N O N

. .

. .. .

. .

. . . .. .. .

23. For the octahedral complexes of Fe3+ in SCN (thiocyanatoS) and in CN ligand

environments, the difference between the spinonly magnetic moments in Bohr magnetons (when approximated to the nearest integer) is

[Atomic number of Fe = 26] 23. [4] Fe+3 i.e. [Ar]18 3d5


16

For weak ligand, 3 22gt eg , i.e. n = 5

For strong ligand, 5 02gt eg , i.e. n = 1

Magnetic moment = n(n 2) B.M.

= 5(5 2) = 35 B.M. for weak ligand.

Magnetic moment for strong ligand = 1(1 2) = 3 B.M.

Difference in magnetic moment = 35 3 = 4

24. Among the triatomic molecules/ions, BeCl2, 3N , N2O, 2NO , O3, SCI2, 2ICl , 3I and

XeF2, the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the dorbital (s) is

[Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54] 24. [4]

1 1 1

Linear (sp hybridisation)

Cl Be Cl N N N N N

25. Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of

H atom is 9, while the degeneracy of the second excited state of H is 25. [3] 26. All the energy released from the reaction X Y, rG

0 = 193 kJ mol1 is used for oxidizing M+ as M+

M3+ + 2e, E0 = 0.25 V. Under standard conditions, the number of moles of M+ oxidized when one mole of X is

converted to Y is [F = 96500 C mol1] 26. [4] G0 = n FE

n = 3193 10

96500 0.25 = 8

No. of moles of M+ oxidized by 1 mol of X = 8

2 = 4

27. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chlorideammonia

complex (which behaves as a strong electrolyte) is 0.0558 C, the number of chloride(s) in the coordination sphere of the complex is

[Kf of water = 1.86 K kg mol1] 27. [1]

Tf = i Kf m

0.0558 = i 1.86 0.01

i = 3

i.e. 2 Cl ions are present outside and one Cl present inside the co-ordination sphere.


17

OCH3

CH3 CH3

H

**

28. The total number of stereoisomers that can exist for M is

28. [2] Compound M have ‘2’ stereo centres.





29. The correct statement (s) about Cr2+ and Mn3+ is (are) [Atomic numbers of Cr = 24 and Mn = 25] (A) Cr2+ is a reducing agent (B) Mn3+

is an oxidizing agent (C) Both Cr2+ and Mn3+

exhibit d4 electronic configuration (D) When Cr2+ is used as a reducing agent, the chromium ion attains d5 electronic

configuration

29. (A), (B), (C) Cr+2 = [Ar]18 3d4 Mn+3 = [Ar]18 3d4 Cr+2 Cr+3 i.e. [Ar]18 3d3 (R.A) Hence, statement (D) is false.

Chromium (II) compounds are reducing agents as they are readily converted into chromium (III) compounds even by the oxygen of the air. Mn+3 + e Mn+2 (O.A) (Most stable)

30. Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process is (are)

(A) Impure Cu strip is used as cathode (B) Acidified aqueous CuSO4 is used as electrolyte (C) Pure Cu deposits at cathode (D) Impurities settle as anodemud


18

30. (B), (C), (D) Impure copper strip is used as anode. 31. Fe3+ is reduced to Fe2+ by using (A) H2O2 in presence of NaOH (B) Na2O2 in water (C) H2O2 in presence of H2SO4 (D) Na2O2 in presence of H2SO4

31. (A), (B) H2O2 oxodizes acidic ferrous sulphate to ferric sulphats. 2 FeSO4 + H2SO4 + H2O2 Fe2 (SO4)3 + 2H2O

2 Na3 Fe(CN)6 + 2 NaOH + H2O2 2 Na4 [Fe(CN)6] + 2H2O + O2 32. The % yield of ammonia as a function of time in the reaction 2 2 3N (g) 3H (g) 2NH (g), H 0

at (P, T1) is given below.

If this reaction is conducted at (P, T2), with T2 > T1, the % yield of ammonia as a function of time is represented by (A) (B) (C) (D)

32. (C) N2(g) + 3H2(g) 2NH3 H < 0 (exothermic).

For exothermic reaction, high temperature favours the reaction in backward direction i.e. yield of reaction decrease.

33. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m

fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are

(A) 1 1

,2 8

(B) 1

1,4

(C) 1 1

,2 2

(D) 1 1

,4 8

T1


19

CH3CH3

H Br2H / Ni

CH3

CH3

H Br

Optically Active

CH3

BrHCH2 2H / Ni

CH3H Br

CH3 Optically Inactive

CH3

BrH

C

CH3

CH22H / Ni

CH3

CH

H Br

CH3

CH3

Optically Active

HBrCH3

CH22H / Ni

Br H

Optically Inactive

33. (A) No. of oxygen atoms per unit cell in ccp = 4 (O2) No. of octahedral voids per unit cell = 4 (Al+3) No. of Tetrahedral voids per unit cell = 8 (Mg+2) Total negative charge due to oxygen atoms = 8 Net charge must be zero. m4(3) + 2n(8) + 4(2) = 0 3m + 4n = 2

(A) 3 4

2 8 = 2 is correct.

(B) 1

3 1 4 4 24

is incorrect.

(C) 1 1 7

3 4 22 2 2

is incorrect.

(D) 1 1

34 8

= 3 2

4 4 =

52

4 is incorrect.

34. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is(are)

(A) (B)

(C) (D) 34. (B) , (D)

(A)

(B)

(C) (D)


20

35. The major product of the following reaction is

(A) (B) (C) (D) 35. (A)

O

CH2

O

CH32(i)KOH,H O

(ii) H ,heat

OH

O

CH

O

CH3

OCH3

O2H O

OH

OHCH3

O

2H O

O

CH3

12

365 4

12345

6

36. In the following reaction, the major product is

(A) (B)

(C) (D)


21

36. (D)

H3 2 2 3 2CH C CH CH CH C CH CH

CH3CH3

CH2

CH2

CH3

Br Br

3 2CH C CH CH

Br

CH3

3 2CH C CH CH Br

CH3

(K.C.P) (T.C.P.)

37. The structure of D-(+)-glucose is The structure of L-()-glucose is (A) (B) (C) (D)

37. (A)

D-Glucose L-Glucose


22

38. The major product of the reaction is (A) (B) (C) (D) 38. (C)

2NaNO , aq. HCl

0 C

3 2CH CH CH CH COOH

CH32NH

3 2CH CH CH CH COOH

CH3 OH




Marking scheme : For each entry in Column I. +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened 0 If none of the bubbles is darkened 1 In all other cases

(P) (Q) (R) (S) (T) (A)

(P) (Q) (R) (S) (T) (B)

(P) (Q) (R) (S) (T) (C)

(P) (Q) (R) (S) (T) (D)


23

39. Match the anionic species given in Column I that are present in the ore(s) given in Column II.

Column I Column II (A) Carbonate (P) Siderite (B) Sulphide (Q) Malachite (C) Hydroxide (R) Bauxite (D) Oxide (S) Calamine (T) Argentite

39. (A) (P), (Q), (S) ; (B) (T) ; (C) (Q) , (R) ; (D) (R)

(P) = Siderite = FeCO3

(Q) = Malachite = CuCO3 . Cu(OH)2

(R) = Bauxite = Al2O3 . 2H2O

(S) = Calamine = ZnCO3

(T) = Argentite = Ag2S 40. Match the thermodynamic processes given under Column I with the expressions given

under Column II.

Column I Column II (A) Freezing of water at 273 K and 1 atm (P) q = 0

(B) Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions

(Q) w = 0

(C) Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container

(R) Ssys < 0

(D) Reversible heating of H2(g) at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm

(S) U = 0

(T) G = 0 40. (A) (R), (T); (B) (P), (Q), (S); (C) (P), (Q), (S); (D) (S), (T)

(A) (R), (T)

H2O () H2O(s)

G = 0 and U = 0, Ssys < 0

(B) (P), (Q), (S)

Free expansion i.e. w = 0, U = 0, q = 0

(C) (P), (Q), (S)

q = 0, U = 0, w = 0

(D) (S), (T)


24

PART III – MATHEMATICS Section I (Maximum Marks : 32)

This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened 0 In all other cases 41. Let the curve C be the mirror image of the parabola y2 = 4x with respect to the line

x + y + 4 = 0. If A and B are the points of intersection of C with the line y = 5, then the distance between A and B is

41. [4] y2 = 4x a = 1 Image of vertex (0, 0) w.r. to x + y + 4 = 0 is (x1, y1)

so, 1x 0

1

= 1y 0

1

=

2 2

0 0 42

1 1

x1 = y1 = 4 (x1, y1) = (4, 4) = A Image of focus S (1, 0) w.r. to x + y + 4 = 0 is (x2, y2)

2x 1

1

= 2y 1

1

=

2 2

1 0 42

1 1

x2 1 = y2 = 5 (x2, y2) (4, 5) Distance between vertex and focus is 1. Also, AB = Latus rectum = 4a = 4.

42. The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

42. [8] Let ‘n’ is the minimum number of times a fair coin is tossed. P (getting at least two heads) = 0.96

2 3 4

2 n 2 3 n 3 4 n 4

C C C1 1 1 1 1 1

n n n2 2 2 2 2 2

+ n

n 0

C1 1

n2 2

≥ 0.96

0 1 2 n

0 n 1 n 1 2 n 2 n 0

C C C C1 1 1 1 1 1 1 1

n n n n2 2 2 2 2 2 2 2

≥

0 1

0 n 1 n 1

C C1 1 1 1

n n2 2 2 2

+ 0.96

n

1 12 2

≥

n n 11 1

n2 2

+ 0.96

1 ≥

n n 11 1

n2 2

+ 0.96

1 n n 1

1 1n

2 2

≥ 0.96

n = 8

x + y + 4 = 0


25

Alternate Method :

n n

n n0 1

1 11 C C 0.96

2 2

n n

1 11 n 0.96

2 2

0.04 n1

n 12

nmax = 8. 43. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a

way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand

consecutively in the queue. Then the value of m

n is

43. [5] n = 6! 5! m = 6! 4! 5 5C4

m

n =

546! 4! 5 C

6! 5!

= 5

44. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are

tangents to the circle (x 3)2 + (y + 2)2 = r2, then the value of r2 is 44. [2] End points of L.R. (a, 2a) (am2, 2am) (1, 2) (m2, 2m) m2 = 1 and m = 1 Normal y = mx 2m m3 y = (1) x 2 (1) (1)3 x + y 3 = 0 As it is normal to (x 3)2 + (y + 2)2 = r2 r distance from (3, 2) = r to x + y 3 = 0

3 2 3

2

= r

2 = r r2 = 2

45. Let f : be a function defined by x , x 2

f x ,0, x 2

where [x] is the greatest integer less than or equal to x. If

22

1

xf xI dx,

2 f x 1

then the

value of (4I 1) is


26

45. [0]

I =

2 2

1

xf x

2 f x 1 dx

I =

0 1 2 22 2 2 2

1 0 1 2

x f x x f x x f x x f x dx dx dx dx

2 f x 1 2 f x 1 2 f x 1 2 f x 1

I = 0 1 2 2

1 0 1 2

x 0 x 0 x.1 dx x 0dx dx dx

2 0 2 1 2 0 2 0

I = 22

11

x4

I = 1

4(2 1)

I = 1

4

4I 1 = 0 46. A cylindrical container is to be made from certain solid material with the following

constraints : It has a fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container.

If the volume of the material used to make the container is minimum when the inner

radius of the container is 10 mm, then the value of V

250 is

46. [4] V = (r 2)2h

Volume of the material used,

Vm = 4h 2r r 22 r

2

Vm = 4h(r 1) + 2 r2

Vm =

2

2

4 r 1 V2 r

r 2

mdV

dr4V

2

4

r 2 r 1 2 r 24 r 0

r 2

Put r = 12 as inner reading is 10.

4V 100 22048 0

10000

48V48

1000 = 0

V

250 = 4

2

r

h


27

47. Let

2x6

2

x

F x 2cos t dt

for all x and 1

f : 0, 0,2

be a continuous function.

For 1

a 0, ,2

if F(a) + 2 is the area of the region bounded by x = 0, y = 0, y = f(x) and

x = a, then f(0) is 47. [3]

F(x) =

2x6

x

(1 cos 2t) dt

=

22 xx66

xx

sin 2tt

2

= 2 21x x sin 2x sin 2x

6 2 3

F(x) = 212x 1 cos 2x x cos 2x

2 3

Given Question,

a

0

f (x) dx = F(a) + 2

= 212a 1 cos 2a 4a 2cos 2a 2

2 3

Diff. w.r.t. a

f(a) = 2 212 4 cos 2a 4a sin 2a 4a 4sin 2a

2 3 3

Put a = 0

f(0) = 1 1

2 4 0 02 2

= 3

48. The number of distinct solutions of the equation 2 4 4 6 65

cos 2x cos x sin x cos x sin x 24

in the interval [0, 2] is 48. [8]

25cos 2

4x + cos4 x + sin4 x + cos6 x + sin6 x = 2

5

4cos2 2x + (cos2 x + sin2 x)2 2cos2 x sin2 x + (cos2 x + sin2 x)3 3cos2 x sin2 x = 2

5

4cos2 2x + 1

2sin 2x 31

2 4 sin2 2x = 2

5

4cos2 2x

5

4sin2 2x = 0

5

4 cos 4x = 0

cos 4x = 0 Number of solution is 8.


28

Section II (Maximum Marks : 40)




49. Let y(x) be a solution of the differential equation (1 + ex)y + yex = 1. If y(0) = 2, then which of the following statements is (are) true?

(A) y(4) = 0 (B) y(2) = 0 (C) y(x) has a critical point in the interval (1, 0) (D) y(x) has no critical point in the interval (1, 0) 49. (A) (C)

(1 + ex) dy

dx

+ y ex = 1

x

x

dy e

dx 1 e

y =

x

1

1 e

I.F. =

x

xe

du 1 ee

= x

elog 1 ee = 1 + ex

So y . (1 + ex) = C + xx

11 e

1 e

du

y (1 + ex) = C + x

y = x

4 x

1 e

y(0) = 2 2 (1 + 1) = C + 0 C = 4 Put x = 4 y(4) = 0

For critical points dy

dx = 0

x x

2x

1 e 1 x 4 e

1 e

= 0

1 3ex x ex = 0 (3 + x) ex = 1 3 + x = ex Graphically one root lies in (1, 0)


29

50. Consider the family of all circles whose centers lie on the straight line y = x. If this family of circles is represented by the differential equation Py + Qy + 1 = 0, where P, Q are

functions of x, y and y (here 2

2

dy d yy ' , y ''

dx dx ), then which of the following statements

is (are) true? (A) P = y + x (B) P = y x (C) P + Q = 1 x + y + y + (y)2 (D) P Q = x + y y (y)2

50. (B) (C) Let circle (x h)2 + (y h)2 = r2

Diff. (x h) + (y h) dy

dx = 0 …(1)

dy dy

x h y hdx dx

= 0

dyx y

dxdy

1dx

= h … (2)

(x h) + (y h) dy

dx = 0

Again diff.

1 + (y h) 22

2

d y dy

dx dx

= 0

1 + 22

2

dyx y d y dydxy

dy dx dx1dx

= 0

1 + 2

2

y x d y dydy dx dx1dx

=0

22

2

d y dy dy dyy x 1 1

dx dx dx dx

= 0

(y x) y+ (y)2 (1 + y) + 1 + y = 0 (y x) y + (y+ (y)2 + 1) y + 1 = 0 P = y x, Q = (y)2 + y + 1 51. Let g: be a differentiable function with g(0) = 0, g(0) = 0 and g(1) 0. Let

f(x) =

xg(x), x 0

| x |

0, x 0

and h(x) = |x|e for all x . Let (f h) (x) denote f(h(x)) and (h f) (x) denote h(f(x)). Then which of the following is (are) true? (A) f is differentiable at x = 0 (B) h is differentiable at x = 0 (C) f h is differentiable at x = 0 (D) h f is differentiable at x = 0


30

51. (A), (D) g(0) = 0, g(0) = 0, g(1) 0

g(x), x 0

f (x) g(x), x 0

0, x 0

For option (A)

R.H.D at x = 0 h 0

f (h) f (0)f (0 ) lim

h

h 0

g(h) g(0)lim g (0) 0

h

L.H.D at x = 0, h 0

f ( h) f (0)f (0 ) lim

h

h 0

g( h) g(0)lim

h

g (0) 0

diff. at x = 0

For (B)

x

x

e , x 0

h(x) e , x 0

1, x 0

t

t 0 t 0

h(t) h(0) e 1h (0 ) lim lim 1

t t

t

t 0 t 0

h( t) h(0) e 1h (0 ) lim lim 1

t t

Not diff. at x = 0 (C) foh f (h(u)) g(h(u) as h(u) > 0

R.H.D t 0

g(h(t)) g(h(0))lim

t

h

t 0

g(e ) g(1)lim g (1) k

t

L.H.D t 0

g(h( t)) g(h(0))lim

t

h

t 0

g(e ) g(1)lim

t

g (1) k Not diff. foh at x = 0 (D) hof = h(f(u))

R.H.D t 0

h(f (t)) h(f (0))lim

t

As g(0) = 0

h 0

g(h) g(0)lim 0

h

t 0

h(g(t)) h(0)lim

t

t 0

h(g(t)) 1lim 0

t


31

L.H.D h 0

h(t( t)) h(t(0))lim

t

h 0

h( g( t)) 1lim 0

t

Diff. at x = 0

52. Let f(x) = sin sin sin x6 2

for all x and g(x) =

2

sin x for all x . Let

(f g) (x) denote f(g(x)) and (g f) (x) denote g(f(x)). Then which of the following is (are) true?

(A) Range of f is 1 1

,2 2

(B) Range of f g is 1 1

,2 2

(C) x 0

f (x)lim

g(x) 6

(D) There is an x such that (g f) (x) = 1

52. (A) (B), (C)

f(x) = sin sin sin x6 2

g(x) = 2

sin x

f (g(x)) = sin sin sin sin x6 2 2

x 0

f xlim

g x =

sin sin sin x sin sin x6 2 6 2

.6sin xsin sin x

26 2

g (f(x)) = sin sin sin sin x2 6 2

1 sin

2 2

≤ g (f(x)) ≤

1 sin

2 2

0.73 ≤ g (f(x)) ≤ 0.73 53. Let PQR be a triangle. Let a = QR , b = RP and c = PQ . If | a | = 12, | b | = 4 3

and b c = 24, then which of the following is (are) true?

(A) 2| c |

| a |2

= 12 (B) 2| c |

| a |2

= 30

(C) | a b c a | = 48 3 (D) a b = 72 53. (A), (C), (D) Given, a = 12

b 4 3, b . c 24

a b c 0

b c a 2

b c a

P

Q R

c b

a


32

2 2b c 2 . b . c a

2c 48

c 4 3

b . c 24

b c cos = 24

cos = 1

2

QPR = 120, PQR = QRP = 30 54. Let X and Y be two arbitrary, 3 3, non-zero, skew-symmetric matrices and Z be an

arbitrary 3 3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

(A) Y3Z4 Z4Y3 (B) X44 + Y44 (C) X4Z3 Z3X4 (D) X23 + Y23 54. (C) (D)

(A) (Y3Z4 Z4Y3)T = (Y3Z4)T (Z4. Y3)T = (Z4)T. (Y3)T (Y3)T.(Z4)T

= Z4. Y3 + Y3Z4 = Y3Z4 Z4Y3 = Y3Z4 Z4Y3 Symmetric

(B) (X44 + Y44)T = (X44)T + (Y44)T = X44 + Y44 Symmetric

(C) (X4 Z3 Z3 X4)T = (X4Z3)T (Z3X4)T = (Z3)T.(X4)T

(X4)T.(Z3)T = (X4Z3

Z3X4) Skew Symmetric

(D) (X23 + Y23)T = X23 Y23 = (X23 + Y23) Skew Symmetric. 55. Which of the following values of satisfy the equation

2 2 2

2 2

2 2 2

(1 ) (1 2 ) (1 3 )

(2 ) (2 2 ) (2 3 )

(3 ) (3 2 ) (3 3 )

= 648?

(A) 4 (B) 9 (C) 9 (D) 4


33

55. (B, C)

2 2 2

2 2 2

2 2 2

1 1 2 1 3

2 2 2 2 3

3 3 2 3 3

= 648

2 2 2

2 2 2

2 2 2

1 2 1 4 4 1 9 6

4 4 4 4 8 4 9 12

9 6 9 4 12 9 9 18

2 2 1

3 3 2

R R R

R R R

= 648α

2 2 21 2 1 4 4 1 9 6

3 2 3 4 3 6

5 2 5 4 5 6

1 1 3

2 2 3

c C C

C C C

= 648α

2 2 24 4 4 9 4

2 2 2

5 2 5 4 5 6

= 648

2 2 24 4 4 9 4

2 1 1 1 648

5 2 5 4 5 6

2

2 2 24 3 8

1 0 0 648

5 2 2 4

2(123 163) = 648 2(43) = 648 83 = 648 (2

81) = 0 = 0, 9, 9 56. In 3, consider the planes P1 : y = 0 and P2 : x + z = 1. Let P3 be a plane, different from P1

and P2, which passes through the intersection of P1 and P2. If the distance of the point (0, 1, 0) from P3 is 1 and the distance of a point (, , ) from P3 is 2, then which of the following relations is (are) true?

(A) 2 + + 2 + 2 = 0 (B) 2 + 2 + 4 = 0 (C) 2 + 2 10 = 0 (D) 2 + 2 8 = 0

56. (B) (D) P1 : y = 0 P2 : x + z = 1 So plane, x + z 1 + y = 0 Given : Distance from (0, 1, 0) from P3 is = 1.

2

0 0 1

1 1

= 1

2 + 1 2 = 2 + 2

= 1

2

So Plane is x + z 1 + 1

y 02

2x y 2z 2 0


34

Given :

2 2

2 2 2

2 1 2

= 2

2α + 2 2 = 6 2α + 2 = 4 or 8 P3 : x + z 1 + y = 0 57. In 3, Let L be a straight line passing through the origin. Suppose that all the points on L

are at a constant distance from the two planes P1 : x + 2y z + 1 = 0 and P2 : 2x y + z 1 = 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane P1. Which of the following points lie(s) on M?

(A) 5 2

0, ,6 3

(B)

1 1 1, ,

6 3 6

(C) 5 1

,6 6

(D) 1 2

, 0,3 3

57. (A) (B) Given x 2y z 1 2x y z 1

1 4 1 4 1 1

x + 2y z + 1 = (2x y + z 1) (+) x 3y + 2z 2 = 0 …(1) () 3x + y = 0 …(2)

Now Let line be x 0 y 0 z 0

m n

…(3)

So, line (3) lie on plane (1) and (2) 3m + 2n = 0

3 + m + 0n = 0

m n

1 3 5

So line is x y zr

1 3 5

Let any point (x, y, z) = (r, 3r, 5r) foot of 1st from (r, 3r, 5r) to plane P1 is x r y 3r z 5r

1 2 1

= r 6r 5r 1

1 4 1

x r y 3r z 5r 1

1 2 1 6

(x, y, z) 1 1 1r , 3r ,5r

6 3 6

For r = 0, option (B)

R = 1

6 option (A)

58. Let P and Q be distinct points on the parabola y2 = 2x such that a circle with PQ as diameter passes through the vertex 0 of the parabola. If P lies in the first quadrant and the area of the

triangle OPQ is 3 2 , then which of the following is (are) the coordinates of P?

(A) (4, 2 2) (B) (9, 3 2) (C) 1 1

,4 2

(D) (1, 2)


35

58. (A) (D)

y2 = 2x a = 1

2

Let P (a 21t , 2at1) =

21

1

t, t

2

and Q (a 22t , 2at2) =

22

2

t, t

2

on parabola.

Equation of circle on PQ as diameter

2 21 2t t

x x2 2

(u t1) (y t2 ) = 0

Given it passes vertex 0 (0, 0)

2 21 2t t

4 + t1t2 = 0

As t1t2 0 t1t2 = 4 (1)

Given OPQ = 3 2

2 21 2 1 21 t t t t

2 2 2

= 3 2

| t1t2 (t1 t2) | = 12 2

t1t2 (t1 t2) = 12 2

(4) (t1 t2) = + 12 2

t1 t2 = 3 2

(+) ()

t1 t2 = 3 2 t1 t2 = 3 2

t1 1

4

t

= 3 2 t1 + 1

4

t = 3 2

21t 3 2 t1 + 4 = 0 2

1t 3 2 t1 + 4 = 0

t1 = 3 2 18 16

2 1

t1 =

3 2 2

2

= 3 2 2

2

= 2 2, 2

= 2 2, 2

Point 21

1

t, t

2

= 4, 2 2 1, 2


36




Marking scheme : For each entry in Column I. +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened 0 If none of the bubbles is darkened 1 In all other cases

59. Column I Column II (A) In 2, if the magnitude of the projection

vector of the vector ˆ î j on ˆ ˆ3 i j is

3 and if = 2 + 3 , then possible value(s) of || is (are)

(P) 1

(B) Let a and b be real numbers such that the function

f(x) = 2

2

3ax 2, x 1

bx a , x 1

is differentiable for all x . Then possible value(s) of a is (are)

(Q) 2

(C) Let 1 be a complex cube root of unity. If (3 3 + 22)4n+3 + (2 + 3 32)4n+3 + (3 + 2 + 32)4n+3 = 0, the possible value(s) of n is (are)

(R) 3

(D) Let the harmonic mean of two positive real numbers a and b be 4. If q is a positive real number such that a, 5, q, b is an arithmetic progression, then the value(s) of |q a| is (are)

(S) 4

(T) 5

(P) (Q) (R) (S) (T) (A)

(P) (Q) (R) (S) (T) (B)

(P) (Q) (R) (S) (T) (C)

(P) (Q) (R) (S) (T) (D)


37

59. A) (P), (B) (P) (Q); (C) (P) (Q) (S) (T); (D) (Q), (T) (A) (P)

2

i j 3 i j

3 1

= 3 Given : = 2 + 3 (2)

3 d + = 2 3 (1) Solving = 2, = 0 (B) (P) (Q)

f(1+) =

h 0

f 1 h f 1lim

h

= 2 2

h 0

b 1 h a b alim

h

= b (1)

f(1) = 2 2

h 0

3a 1 h 2 b alim

h

=

2 2

h 0

3a 1 2h h 2 b alim

h

= 2 2

h 0

3ah 6ah 3a 2 b alim

h

= 6a (2) [as differentiable 3a 2 b a2 = 0 a2 + 3a + 2 + b = 0] (4) f (1+) = f (1) b = 6a (3) From (3) and (4) a2 + 3a + 2 6a = 0 a2 3a + 2 = 0 a = 1, 2 (C) (P) (Q) (S) (T)

(3 3w + 2w2)4n + 3 + 4n 3 4n 32 2

2

3 3w 2w 3 3w 2w

ww

= 0

(3 3w + 2w2)4n+3 + 4n 3 4n 3

2

1 11

ww

= 0

1 + (w)4n+3 + (w2)4n+3 = 0 4n + 3 is not multiple of 3 (D) (Q), (T)

2ab

a b = 4

ab

a b = 2

a(15 2a)2

a 15 2a

5a ,6

2 …(1)

a, 5, q, b in AOP. 10 = a + q; 2q = 5 + b a + b + q + 5 = w + 2q b 5 = q a …(2) from (1) and (2) q a = 5 or 2


38

60. Column I Column II (A) In a triangle XYZ, let a, b and c be the

lengths of the sides opposite to the angles X, Y and Z, respectively. If 2(a2 b2) = c2 and

= sin (X Y)

sin Z

, then possible values of n

for which cos(n) = 0 is (are)

(P) 1

(B) In a triangle XYZ, let a, b and c be the lengths of the sides opposite to the angles X, Y and Z, respectively. If 1 + cos 2X 2cos 2Y = 2 sin X sin y, then possible value(s) of a

b is (are)

(Q) 2

(C) If 2, let ˆ ˆ3i j , ˆ î 3 j and ˆ î (1 ) j be the position vectors of X, Y and Z with respect to the origin O, respectively. If the distance of Z from the bisector of the acute

angle of OX with OY is 3

2, then

possible value(s) of | | is (are)

(R) 3

(D) Suppose that F() denotes the area of the region bounded by x = 0, x = 2, y2 = 4x and y = |x 1| + |x 2| + ax, where {0,

1}. Then the value(s) of F() + 8

23

, when

= 0 and = 1, is (are)

(S) 5

(T) 6 60. (A) (P), (R), (S); (B) (P); (C) (P), (Q); (D) (S), (T)

(A) (P), (R), (S); 2(a2 b2) = c2

2 2 22(sin X sin Y) sin (Z)

22sin(X Y)sin(X Y) sin (Z)

sin(X Y) 1

sin(Z) 2

( sin(X Y) sin Z)

1

2

n

cos 02

for n = 1, 3 and 5

(B) (P)

1 cos(2X) 2cos(2Y) 2sin(X)sin(Y)

2 21 1 2sin (X) 2(1 2sin (Y) 2sin(X)sin(Y)

2 22sin (Y) sin (X) sin(X)sin(Y)


39

2

2

sin (X) sin(X)2

sin(Y)sin (Y)

Let sin(X) a

ksin(Y) b

22 k k k2 + k 2 = 0 (k + 2) (k 1) = 0 k = 1, 2

(C) (P), (Q)

Z lies on the line x + y = 1 Angle bisector is x y = 0 Let z (, 1 )

Then (1 ) 3

2 1 32 2

= 2, 1

1,2

(D) (S), (T) For = 0, y = 3

2

0

8 2f (0) 3 2 x dx 6

3

8 2

f (0) 63

For = 1, y x 1 x 2 x 1 2

0 1

f (1) 3 x 2 x x 1 2 xdx

8 2 8 25 f (1) 5

3 3

y 1, 3

x 3, 1

3 y2 = 4x

3

0 1 2

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IIT - JEE 2015 (Advanced) - Vidyalankarvidyalankar.org/file/JEE-Advanced-2015-Solution-Paper-I.pdf · (4) Vidyalankar : IIT JEE 2015 Advanced : Question Paper & Solution 4 6. A Young’s